1. Suppose you are using polar coordinates [tex]r,\theta[/tex], what sort of region is described by the inequalities:
[tex]R\leq{r}\leq{R}+\bigtriangleup{R},\theta_{0}\leq\theta\leq\theta_{0}+\bigtriangleup{\theta}[/tex]

You ought to see that this represent a specific portion of a section of a circle.

2. Now, if you let [tex]\bigtriangleup{R},\bigtriangleup\theta[/tex] tend towards zero, you may approximate this region as a rectangle (or trapezoid, if you like that better). What should the (tiny!) area of this figure be?

imagine the top half of a circle. the origin lies along the bottom of the semicircle, and in the middle. y axis up, and x axis to the right and left. i think theta can only go from 0 to 180 degrees since it is a semi circle. Y = d(theta) R squared
R = radius, integrate from 0 to R

Sure I have, but you are so vague and obtuse that I don't think you understand what you write yourself.
So, again:
show that you have at least a minimum of mathematical skill and understanding.
I have already shown you how you can do this problem, but it doesn't seem you have the competence to follow a perfectly clear line of reasoning.

Why can you write this? Where did it come from? You cannot just drop the dx and it is not automaticlly set to 4 or any other value.

One of the first things you need to do is carefully define your rectangle in terms of your coordinate system.

Suppose one corner is at the origin, x=y=0 and the diagonal corner is at x=L, y=W. To find the area we must define our integral:
[tex] \int dA = \int^{y=W}_{y=0} \int^{x=L}_{x=0} dx dy [/tex]

which lets me integrate along y axis,

Click to expand...

Can you complete the above integral?

now for a semicircle, how would i do that if dA = r dr d(theta)

Click to expand...

Ok, now lets set up the integral for the area of a semi circle.
Once again we need to define the circle in terms of the coordinate system.

If the center of the circle is at the origin and the radius is R.

So to get the Area of the circle (or portion there of) the radius will take on values between 0 and R, while the angle [itex]\Theta[/itex] varies from 0 to [itex] \pi[/itex] (for a semi circle). Now lets set up the integral

imagine the top half of a circle. the origin lies along the bottom of the semicircle, and in the middle. y axis up, and x axis to the right and left. i think theta can only go from 0 to 180 degrees since it is a semi circle. Y = d(theta) R squared
R = radius, integrate from 0 to R

Click to expand...

You haven't stated a problem here! What do you want to integrate from 0 to R? What function are you integrating?

If you just want the area, look at the problem you gave:

in a rectangles case, dA = dX dY

if i have the rectangles width and it is 4 m, then dA = 4dY

Click to expand...

(And arildno did not say he had never seen that before, he was trying to get you to think about why you think that was true.)

Suppose your semi-circles radius is R. Then dA= r dr dθ. In a sense, dr here, as well as r, is r: along each radius we go from r= 0 to R so the "change in r", dr, is R.

Okay, then r dr dθ becomes dA= R2 dθ, just like your dX dY became 4dy, the area of the semi-circle is
[tex]\int_{\theta=0}^{\pi} R^2 d\theta= \pi R^2[/tex], exactly the right answer.

(θ does not go from 0 to 180! As I am sure you learned in when you were learning the derivatives of sine and cosine, in calculus, all angles are in radians.)

(Oh, and by the way, why was this posted under "differential equations". Was this part of a differential equations problem?)